DSP1

Digital Signal Processing. Slide 1.1

Module 1

Sampling and Aliasing, System Functionsand z-transforms


Contents

Sampling and Aliasing Concept of sampling a continuous-time signal Periodic nature of discrete-time signals in the frequency domain, the effect of

aliasing and Nyquist sampling criterion Reconstruction of continuous-time signals from their samples

System Functions Transfer function Frequency response Example

z-transforms Definition and properties Inverse z-transform Causal and anticausal systems Stability


Sampling

Analog signal Sampling function sT(t) is a sequence of impulses

x ta ( ) x nTa ( )

s tT ( )-0.5

1

0 10 20 30 40 50-1

0

0.5

x t( )

x n T( )s tT ( )

t

s t t mTm

T ( ) ( )= =

Sampling period T

x ta ( )


Symbols

continuous time signal samples of continuous time signal discrete-time signal frequency digital frequency FT of DFT of

x ta ( )x nTa ( )x n( )

x ta ( )x n( ) ( )X e j

)(aX


Analysis of Sampled Signal

Spectrum of discrete-time signal = spectrum of continuous-time signal + images at multiples of 2 From IFT:

From IDFT:

Combining these:

x n X e e dj j n( ) ( )=12

X eT

X jT

rT

ja

r( ) = +

=

1 2

==

dejXnTxnx nTjaa )(21)()(


Proof write expression for as sum of integrals over intervals of length

change of variable: replace with

use integer, reverse order of sum and integration and use

x nTa ( )2 T

=

+

=r

Tr

Tr

nTja dejXnx

)12(

)12(

)( 21)(

Tr2'+

=

+=r

T

T

nrnTja deeT

rjXnx '2' 21)( 2'

),( 12 nre rnj =

de

Tr

TjX

Tnx nj

ra

=

+= 2121)(

T='


Note that this is in the form of an IDFT since Hence

=

+=r

aj

Tr

TjX

TeX 21)(

de

Tr

TjX

Tnx nj

ra

=

+= 2121)(

[1]


Points to note: spectrum of x(n) is periodic in with period 2 if Xa() is not bandlimited to then information in the signal is lost

when sampled due to overlapping spectral images - this effect is called aliasing

if Xa(j) is bandlimited to then the original continuous-time signal can be perfectly reconstructed from its discrete-time samples

this is known as the Nyquist Sampling Criterion is the analog frequency,

is the digital frequency

s2

s2

s 2

0 < < = 2f = = =T fT f

f s2 2

22

X e j( )

s 2

1 1/T

)(aX


Examples of signal spectra after sampling 1) sinusoidal signal at 1 kHz, sampling frequency = 8 kHz

2) sinusoidal signal at 5.5125 kHz, sampling frequency = 44.1 kHz

22etc

X e j( )

22etc

X e j( )


3) sinusoidal signal at 1 kHz, sampling frequency = 1.1429 kHz

22 Alia

sedAlia

sedAlia

sedAlia

sed X ej( )


Sampling in time domain => periodicity in frequency

Sampling in frequency domain => periodicity in time


Signal Reconstruction

A continuous-time signal can be reconstructed from its samples as

where

and corresponds to a lowpass filter with cut-off at the Nyquist frequency.

)(txa{ })(nTxa=

=n

aa nTtgnTxtx )()()(

( )Tt

Tttg sin)( =


Proof write the IFT expression for for the range

or equivalently

from [1] we know that, in the range

giving

x ta ( ) T T

x t TX e e daj T j t

T

T( ) ( )

/

/=

12

=T

T

tjaa deXtx

/

/

)(21)(

)(1)()( == aTjj XTeXeX


Proof (continued) using the DTFT relation (described later)

write

change order of summation and integration

X e x nT ej T aj nT

n( ) ( ) =

=

x t T x nT e e da aj nT

n

j t

T

T( ) ( )

/

/=

=

2

( )( )nTt

T

nTtTnTx

deTnTxtx

na

T

T

nTtj

naa

=

=

=

=

sin)(

2)()(

/

/

)(


Proof (continued) This operation can be recognized as the convolution of with the

sinc function

This convolution represents filtering with an ideal lowpass filter with a cut-off frequency of

the Nyquist frequency

)(nTxa

( )Tt

Ttsin

=

Reading: Proakis: Chapter 1, especially 1.4.1 to 1.4.7


System Functions

Transfer function For a continuous-time system H(s) with input X(s) and output Y(s),

its transfer function is defined as

For a discrete-time system H(z) with input X(z) and output Y(z),

its transfer function is defined as

H(.), Y(.) and X(.) are polynomials in (.)

H s Y sX s

( ) ( )( )

=

H z Y zX z

( ) ( )( )

=


Frequency response For continuous-time systems, use and investigate the

function H(s) as a function of frequency only, i.e. write

For discrete-time systems, use and investigate the function H(z) as a function of frequency only, i.e. write

s j= + s j=

z e s jsT= = +, z e j=

j

Re

Imz = 1 s-plane z-plane

s j=


Example: Transfer Function

zeros at z = 0.3+j0.3 and z = 0.3-j0.3 poles at z = -0.9 and z = 0.7

Can be written in terms of as

Difference Equation:

-1 -0.5 0 0.5 1

-1

-0.5

0

0.5

1

Real part

I

m

a

g

i

n

a

r

y

p

a

r

t

-1 -0.5 0 0.5 1

-1

-0.5

0

0.5

1

Real part

I

m

a

g

i

n

a

r

y

p

a

r

t

y n x n x n x n y n y n( ) ( ) . ( ) . ( ) . ( ) . ( )= + + 0 6 1 018 2 0 2 1 0 63 2

z-plane

H z z zz z

( ) . .. .

= ++ 2

20 6 0180 2 0 63

1z21

21

63.02.0118.06.01)(

++=

zzzzzH


Frequency Response

set

plot magnitude and phase

normally plot for normalized such that

0 0.2 0.4 0.6 0.8 1-500

50100

Normalized frequency (Nyquist == 1)

P

h

a

s

e

(

d

e

g

r

e

e

s

)

0 0.2 0.4 0.6 0.8 1-100

102030

M

a

g

n

i

t

u

d

e

(

d

B

) Frequency Responsez e j=

0 <


Notation

Z-transform denoted by

relationship indicated by

{ })()( nxZzX

)()( zXnx z


z-transform is an infinite power series only exists for particular values of z for which the series converges these are the values of z for which X(z) has a finite value

need to specify Region Of Convergence (ROC) when referring to z-transform

Region of ConvergenceX z x n z n

n( ) ( )=

=


Examples

Finite Duration Sequences

{ } 4321 85321)( 8 5, 3, 2, 1,)( ++++== zzzzzXnx

{ }

++++==

8532)( 8 5, 3, 2, 1,)( 2112 zzzzzXnx

1)( )()( == zXnnx

ROC: zz and 0

ROC: 0z

ROC: zDigital Signal Processing. Slide 1.24

Infinite Duration Sequences

Sequences like are called right-sided sequences

x n Au n( ) ( )=

( )X z Au n z

A z z z

Az

n

n( ) ( )=

= + + + +

=

=

1

1

1 2 3

1

"

z


Another Example

The region of convergence is

orsince

z planeROC

z a=

e j = 1

1)()( jnneanAunx =

( )1

0

1

1

1

1

1

)()(

=

=

=

=

=

zaeA

zaeA

zeanAuzX

j

n

nj

n

njnn

az

zae j

>


Example Find the z-transform of the following function

Write

Using the shift and linearity properties we obtain

The ROC is

0 n1 2 3 N-1-1-2-3

1

p n( )

p n u n u n N( ) ( ) ( )=

P zz

zz

zz

N N( ) = =

11 1

111 1 1

z


Inverse z-transform by inspection Given a z-domain expression as a power series

use

to write

X z z z( ) = + + 1 2 31 2

{ }Z A n m Az m ( ) =

{ }3 ,2 ,1)2(3)1(2)()(

=++= nnnnx


Inverse z-transform by long division Given a z-domain expression as a ratio of polynomials, the first few terms

of the sequence can be found by long division. Start by converting ratio of polynomials to power series, then use

inspection E.g.

and hence

X z z zz z

( ) . ..

= + +0 5 0 5

0 5

2

2

. + .

.

0 etc

-10 5 10 0 75

0 5 0 5 0 5

0 5 0 5 0 250 10 0 25

10 100 0 50

0 75 0 50

2

2 2

2

1

1

z z

z z z z

z zz

z z

z

+ + + +

++

++

. ...

. ) . .

. .. .

. . .

. .

{ }

,75.0 ,1 ,5.0)2(75.0)1(1)(5.0)(

=+++= nnnnx


Inverse z-transform by partial fractions and table look-up Use tables of standard transform pairs Use partial fraction expansion to re-write problem in terms of standard

transform pairs E.g.

Use PFE to write

Use standard transform pair

to give

X z zz

( ).

= 4

0 25

2

2

x n u n u nn n( ) ( . ) ( ) ( . ) ( )= + 2 05 2 05

X z zz

zz

( ). .

= + +2

052

05

{ }Z Aa u n Azz an ( ) =


Inverse z-transform by the inversion formula The inverse z-transform is given by

This can be solved using the residue theorem

Express as

which has s poles at

Then

x nj

X z z dznC

( ) ( )= 12 1

( )x n j X z z dz X z znC n( ) ( ) ( )= = 12 1 1 residues of at the poles inside contour CX z zn( ) 1 X z z z

z zn

s( )( )

( ) =

1

0

z z= 0

Re [ ( ) ]( )!

( )s X z z z zs

d zdz

ns

sz z

== =

10

1

11

10

at


Example Find the inverse z-transform of

Write

C is a circular contour of radius greater than a.

Comparing with the form

gives , and .

For the only pole of is at with a residue of

z a0 =s = 1 ( )z zn=

n 0 z a=X z zn( ) 1

azaz

zX >= for 11)( 1

==

C

n

C

n

dzaz

zj

dzaz

zj

nx 21

121)( 1

1

sn

zzzzzX

)()()(0

1

=

na


For there is a multiple order pole at For n= -1

residue of pole at origin is residue of pole at is

For n=-2 residue of pole at origin is residue of pole at is

etc.

Therefore

n < 0 z = 0

az = a 1 cancel

Res 122

z z aa

z a( )

==

z a=

Res 120

2

z z aa

z( )

= =

cancel

1 a

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